Python flatten list (but not all the way)

Question

I have a list of lists of lists...

A = [ [[1,3]], [[3,5], [4,4], [[5,3]]] ]

the following function outputs [1, 3, 3, 5, 4, 4, 5, 3]

def flatten(a):
    b = []
    for c in a:
        if isinstance(c, list):
            b.extend(flatten(c))
        else:
            b.append(c)
    return b

However, I want to stop flattening on the last level so that I get [ [1,3], [3,5], [4,4], [5,3] ]

This would require a scan of the list to see if there are no lists contained. — Martijn Pieters, Oct 27 '13 at 11:11
@MartijnPieters Yea, I get the above, but the problem is that last element is a list within a list. — Games Brainiac, Oct 27 '13 at 11:12
@GamesBrainiac: see my answer; just test for lists before flattening. — Martijn Pieters, Oct 27 '13 at 11:14
@MartijnPieters I did that with list comprehension. But yours is generic :) — thefourtheye, Oct 27 '13 at 11:15
You can use this [post](http://stackoverflow.com/a/5251706/1982962) that's the missing part, to check if your list has lists inside it before making the recursion. — Kobi K, Oct 27 '13 at 11:16

score 4 · Accepted Answer · answered Oct 27 '13 at 11:11

You could test for contained lists before flattening:

def flatten(a):
    b = []
    for c in a:
        if isinstance(c, list) and any(isinstance(i, list) for i in c):
            b.extend(flatten(c))
        else:
            b.append(c)
    return b

Demo:

>>> def flatten(a):
...     b = []
...     for c in a:
...         if isinstance(c, list) and any(isinstance(i, list) for i in c):
...             b.extend(flatten(c))
...         else:
...             b.append(c)
...     return b
... 
>>> A = [ [[1,3]], [[3,5], [4,4], [[5,3]]] ]
>>> flatten(A)
[[1, 3], [3, 5], [4, 4], [5, 3]]

This tries to be as efficient as can be under the circumstances; any() only needs to test until a list is found, not all elements.

Beautifully done, in the end I did the same thing, but I used `chain.from_iterable`. — Games Brainiac, Oct 27 '13 at 11:22

thefourtheye · Answer 2 · 2013-10-27T13:01:35.410

2

A = [ [[1,3]], [[3,5], [4,4], [[5,3]]] ]

print [child[0] if isinstance(child[0], list) else child for item in A for child in item]

Output

[[1, 3], [3, 5], [4, 4], [5, 3]]

Note: This solution is strictly for this problem only. This is not a generic list flattening solution.

The same idea, with itertools.chain

from itertools import chain
print [item[0] if isinstance(child[0], list) else item for item in chain(*A)]

Output

[[1, 3], [3, 5], [4, 4], [5, 3]]

edited Oct 27 '13 at 13:01

answered Oct 27 '13 at 11:13

thefourtheye

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Use `isinstance()` to test for types; `type(...) == typeobj` is very restrictive. Even if `isinstance()` was not available, you should use `is typeobj` anyway. – Martijn Pieters Oct 27 '13 at 12:41
@MartijnPieters Why is it restrictive? Could you please xplain? – thefourtheye Oct 27 '13 at 12:59
`isinstance(ob, list)` is `True` for subclasses too, `type(ob) is list` only allows for list objects, not subclasses. – Martijn Pieters Oct 27 '13 at 13:00
@MartijnPieters Oh. Thanks. Updated :) Not a single day passes by without learning something from you. Thanks a lot :) – thefourtheye Oct 27 '13 at 13:02

Python flatten list (but not all the way)

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